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3.12 Tensile testing is not appropriate for hard brittle materials such as ceramics. What is the test
commonly used to determine the strength properties of such materials?
Answer. A three-point bending test is commonly used to test the strength of brittle materials. The test
provides a measure called the transverse rupture strength for these materials.
3.13 How is the shear modulus of elasticity G related to the tensile modulus of elasticity E, on average?
Answer. G = 0.4 E, on average.
3.14 How is shear strength S related to tensile strength TS, on average?
Answer. S = 0.7 TS, on average.
3.15 What is hardness and how is it generally tested?
Answer. Hardness is defined as the resistance to indentation of a material. It is tested by pressing a
hard object (sphere, diamond point) into the test material and measuring the size (depth, area) of the
indentation.
3.16 Why are different hardness tests and scales required?
Answer. Different hardness tests and scales are required because different materials possess widely
differing hardnesses. A test whose measuring range is suited to very hard materials is not sensitive for
testing very soft materials.
3.17 Define the recrystallization temperature for a metal.
Answer. The recrystallization temperature is the temperature at which a metal recrystallizes (forms
new grains) rather than work hardens when deformed.
3.18 Define viscosity of a fluid.
Answer. Viscosity is the resistance to flow of a fluid material; the thicker the fluid, the greater the
viscosity.
3.19 What is the defining characteristic of a Newtonian fluid?
Answer. A Newtonian fluid is one for which viscosity is a constant property at a given temperature.
Most liquids (water, oils) are Newtonian fluids.
3.20 What is viscoelasticity, as a material property?
Answer. Viscoelasticity refers to the property most commonly exhibited by polymers that defines the
strain of the material as a function of stress and temperature over time. It is a combination of viscosity
and elasticity.
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3.2 Which one of the following is the correct definition of ultimate tensile strength, as derived from the
results of a tensile test on a metal specimen: (a) the stress encountered when the stress-strain curve
transforms from elastic to plastic behavior, (b) the maximum load divided by the final area of the
specimen, (c) the maximum load divided by the original area of the specimen, or (d) the stress
observed when the specimen finally fails?
Answer. (c).
3.3 If stress values were measured during a tensile test, which of the following would have the higher
value: (a) engineering stress or (b) true stress?
Answer. (b).
3.4 If strain measurements were made during a tensile test, which of the following would have the higher
value: (a) engineering stain, or (b) true strain?
Answer. (a).
3.5 The plastic region of the stress-strain curve for a metal is characterized by a proportional relationship
between stress and strain: (a) true or (b) false?
Answer. (b). It is the elastic region that is characterized by a proportional relationship between stress
and strain. The plastic region is characterized by a power function - the flow curve.
3.6 Which one of the following types of stress strain relationship best describes the behavior of brittle
materials such as ceramics and thermosetting plastics: (a) elastic and perfectly plastic, (b) elastic and
strain hardening, (c) perfectly elastic, or (d) none of the above?
Answer. (c).
3.7 Which one of the following types of stress strain relationship best describes the behavior of most
metals at room temperature: (a) elastic and perfectly plastic, (b) elastic and strain hardening, (c)
perfectly elastic, or (d) none of the above?
Answer. (b).
3.8 Which one of the following types of stress strain relationship best describes the behavior of metals at
temperatures above their respective recrystallization points: (a) elastic and perfectly plastic, (b) elastic
and strain hardening, (c) perfectly elastic, or (d) none of the above?
Answer. (a).
3.9 Which one of the following materials has the highest modulus of elasticity: (a) aluminum, (b)
diamond, (c) steel, (d) titanium, or (e) tungsten?
Answer. (b).
3.10 The shear strength of a metal is usually (a) greater than or (b) less than its tensile strength?
Answer. (b).
3.11 Most hardness tests involve pressing a hard object into the surface of a test specimen and measuring
the indentation (or its effect) that results: (a) true or (b) false?
Answer. (a).
3.12 Which one of the following materials has the highest hardness: (a) alumina ceramic, (b) gray cast
iron, (c) hardened tool steel, (d) high carbon steel, or (e) polystyrene?
Answer. (a).
3.13 Viscosity can be defined as the ease with which a fluid flows: (a) true or (b) false?
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Problems
Strength and Ductility in Tension
3.1 A tensile test uses a test specimen that has a gage length of 50 mm and an area = 200 mm2. During
the test the specimen yields under a load of 98,000 N. The corresponding gage length = 50.23 mm.
This is the 0.2 percent yield point. The maximum load = 168,000 N is reached at a gage length = 64.2
mm. Determine (a) yield strength, (b) modulus of elasticity, and (c) tensile strength.
Solution: (a) Y = 98,000/200 = 490 MPa.
(b) σ = E e
Subtracting the 0.2% offset, e = (50.23 - 50.0)/50.0 - 0.002 = 0.0026
E = σ/e = 490/0.0026 = 188.5 x 103 MPa.
(c) TS = 168,000/200 = 840 MPa.
3.2 In Problem 3.1, fracture occurs at a gage length of 67.3 mm. (a) Determine the percent elongation.
(b) If the specimen necked to an area = 92 mm2, determine the percent reduction in area.
Solution: (a) EL = (67.3 - 50)/50 = 17.3/50 = 0.346 = 34.6%
(b) AR = (200 - 92)/200 = 0.54 = 54%
3.3 A test specimen in a tensile test has a gage length of 2.0 in and an area = 0.5 in2. During the test the
specimen yields under a load of 32,000 lb. The corresponding gage length = 2.0083 in. This is the 0.2
percent yield point. The maximum load = 60,000 lb is reached at a gage length = 2.60 in. Determine
(a) yield strength, (b) modulus of elasticity, and (c) tensile strength.
Solution: (a) Y = 32,000/0.5 = 64,000 lb/in2
(b) σ = E e
Subtracting the 0.2% offset, e = (2.0083 - 2.0)/2.0 - 0.002 = 0.00215
E = σ/e = 64,000/0.00215 = 29.77 x 106 lb/in2
(c) TS = 60,000/0.5 = 120,000 lb/in2
3.4 In Problem 3.3, fracture occurs at a gage length of 2.92 in. (a) Determine the percent elongation. (b)
If the specimen necked to an area = 0.25 in2, determine the percent reduction in area.
Solution: (a) EL = (2.92 - 2.0)/2.0 = 0.92/2.0 = 0.46 = 46%
(b) AR = (0.5 - 0.25)/0.5 = 0.50 = 50%
3.5 During a tensile test in which the starting gage length = 125.0 mm and the cross- sectional area = 62.5
mm2, the following force and gage length data are collected (1) 17,793 N at 125.23 mm, (2) 23,042 N
at 131.25 mm, (3) 27,579 N at 140.05 mm, (4) 28, 913 N at 147.01 mm, (5) 27,578 N at 153.00 mm,
and (6) 20,462 N at 160.10 mm. The maximum load is 28,913 N and the final data point occurred
immediately prior to failure. (a) Plot the engineering stress strain curve. Determine: (b) yield strength,
(c) modulus of elasticity, (d) tensile strength.
Solution: (a) Student exercise.
(b) From the plot, Y = 310.27 MPa.
(c) First data point is prior to yielding.
Strain e = (125.23 - 125)/125 = 0.00184, E = 310.27/0.00184 = 168,625 MPa.
(d) From the plot, TS = 426.6 MPa.
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Flow Curve
3.6 In Problem 3.5, determine the strength coefficient and the strain-hardening exponent in the flow
curve equation. Be sure not to use data after the point at which necking occurred.
Solution: Starting volume of test specimen V = 125(62.5) = 7812.5 mm3.
Select two data points: (1) F = 23042 N and L = 131.25 mm; (2) F = 28913 N and L = 147.01 mm.
(1) A = V/L = 7812.5/131.25 = 59.524 mm2.
Stress σ = 23042/59.524 = 387.1 MPa. Strain ε = ln(131.25/125) = 0.0488
(2) A = 7812.5/147.01 = 53.143 mm2.
Stress σ = 28913/53.143 = 544.1 MPa. Strain ε = ln(147.01/125) = 0.1622
Substituting these values into the flow curve equation, we have
(1) 387.1 = K(0.0488)n and (2) 544.1 = K(0.1622)n
544.1/387.1 = (0.1622/0.0488)n
1.4056 = (3.3238)n
ln(1.4056) = n ln(3.3238) 0.3405 = 1.2011 n n = 0.283
Substituting this value with the data back into the flow curve equation, we obtain the value of the
strength coefficient K:
K = 387.1/(0.0488).283 = 909.9 MPa
K = 544.1/(0.1622).283 = 910.4 MPa Use average K = 910.2 MPa
The flow curve equation is: σ = 910.2 ε0.283
3.7 In a tensile test on a metal specimen, true strain = 0.08 at a stress = 265 MPa. When the true stress =
325 MPa, the true strain = 0.27. Determine the strength coefficient and the strain-hardening exponent
in the flow curve equation.
Solution: (1) 265 = K(0.08)n and (2) 325 = K(0.27)n
325/265 = (0.27/0.08)n 1.2264 = (3.375)n
n ln(3.375) = ln(1.2264) 1.2164 n = 0.2041 n = 0.1678
Substituting this value with the data back into the flow curve equation, we obtain the value of the
strength coefficient K:
(1) K = 265/(0.08).1678 = 404.85 MPa
(2) K = 325/(0.27).1678 = 404.85 MPa
The flow curve equation is: σ = 404.85 ε0.1678
3.8 During a tensile test, a metal has a true strain = 0.10 at a true stress = 37,000 lb/in2. Later, at a true
stress = 55,000 lb/in2, the true strain = 0.25. Determine the strength coefficient and the strain-
hardening exponent in the flow curve equation.
Solution: (1) 37,000 = K(0.10)n and (2) 55,000 = K(0.25)n
55,000/37,000 = (0.25/0.10)n 1.4865 = (2.5)n
n ln(2.5) = ln(1.4865) 0.9163 n = 0.3964 n = 0.4326
Substituting this value with the data back into the flow curve equation, we obtain the value of the
strength coefficient K:
(1) K = 37,000/(0.10)0.4326 = 100,191 lb/in2
(2) K = 55,000/(0.25)0.4326 = 100,191 lb/in2
The flow curve equation is: σ = 100,191 ε0.4326
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3.9 In a tensile test a metal begins to neck at a true strain = 0.28 with a corresponding true stress = 345.0
MPa. Without knowing any more about the test, can you estimate the strength coefficient and the
strain-hardening exponent in the flow curve equation?
Solution: If we assume that n = ε when necking starts, then n = 0.28.
Using this value in the flow curve equation, we have K = 345/(0.28).28 = 492.7 MPa
The flow curve equation is: σ = 492.7 ε0.28
3.10 A tensile test for a certain metal provides flow curve parameters: strain-hardening exponent is 0.3 and
strength coefficient is 600 MPa. Determine (a) the flow stress at a true strain = 1.0 and (b) true strain
at a flow stress = 600 MPa.
Solution: (a) Yf = 600(1.0).3 = 600 MPa
(b) ε = (600/600)1/.3 = (1.0)3.33 = 1.00
3.11 The flow curve for a certain metal has parameters: strain-hardening exponent is = 0.22 and strength
coefficient is 54,000 lb/in2. Determine: (a) the flow stress at a true strain = 0.45, and (b) the true strain
at a flow stress = 40,000 lb/in2.
Solution: (a) Yf = 54,000(0.45).22 = 45,300 lb/in2
(b) ε = (40,000/54,000)1/.22 = (0.7407)4.545 = 0.256
3.12 A metal is deformed in a tension test into its plastic region. The starting specimen had a gage length =
2.0 in and an area = 0.50 in2. At one point in the tensile test, the gage length = 2.5 in and the
corresponding engineering stress = 24,000 lb/in2; and at another point in the test prior to necking, the
gage length = 3.2 in and the corresponding engineering stress = 28,000 lb/in2. Determine the strength
coefficient and the strain-hardening exponent for this metal.
Solution: Starting volume V = LoAo = 2.0(0.5) = 1.0 in3
(1) A = V/L = 1.0/2.5 = 0.4 in2
So, true stress σ = 24,000(0.5)/0.4 = 31,250 lb/in2 and ε = ln(2.5/2.0) = 0.223
(2) A = 1.0/3.2 = 0.3125 in2
So, true stress σ = 28,000(0.5)/0.3125 = 44,800 lb/in2 and ε = ln(3.2/2.0) = 0.470
These are two data points with which to determine the parameters of the flow curve equation.
(1) 31,250 = K(0.223)n and (2) 44,800 = K(0.470)n
44,800/31,250 = (0.470/0.223)n
1.4336 = (2.1076)n
ln(1.4336) = n ln(2.1076)
.3602 = .7455 n n = 0.483
(1) K = 31,250/(0.223).483 = 64,513 lb/in2
(2) K = 44,800/(0.470).483 = 64,516 lb/in2 Use average K = 64,515 lb/in2
The flow curve equation is: σ = 64,515 ε0.483
3.13 A tensile test specimen has a starting gage length = 75.0 mm. It is elongated during the test to a
length = 110.0 mm before necking occurs. Determine (a) the engineering strain and (b) the true strain.
(c) Compute and sum the engineering strains as the specimen elongates from: (1) 75.0 to 80.0 mm,
(2) 80.0 to 85.0 mm, (3) 85.0 to 90.0 mm, (4) 90.0 to 95.0 mm, (5) 95.0 to 100.0 mm, (6) 100.0 to
105.0 mm, and (7) 105.0 to 110.0 mm. (d) Is the result closer to the answer to part (a) or part (b)?
Does this help to show what is meant by the term true strain?
Solution: (a) Engineering strain e = (110 - 75)/75 = 35/75 = 0.4667
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Solution: Tensile strength occurs at maximum value of load. Necking begins immediately
thereafter. At necking, n = ε. Therefore, σ = 551.6(0.4)0.4 = 382.3 MPa. This is a true stress.
TS is defined as an engineering stress. From Problem 3.15, we know that ε = 2 ln(Do/D). Therefore,
0.4 = 2 ln(Do/D)
ln(Do/D) = .4/2 = 0.2
Do/D = exp(0.2) = 1.221
Area ratio = (Do/D)2 = (1.221)2 = 1.4918
The ratio between true stress and engineering stress would be the same ratio.
Therefore, TS = 1.4918(382.3) = 570.3 MPa
3.18 A copper wire of diameter 0.80 mm fails at an engineering stress = 248.2 MPa. Its ductility is
measured as 75% reduction of area. Determine the true stress and true strain at failure.
Solution: Area reduction AR = (Ao - Af)/Ao = 0.75
Ao - Af = 0.75 Ao
Ao - 0.75Ao = 0.25 Ao = Af
If engineering stress = 248.2 MPa, then true stress σ = 248.2/0.25 = 992.8 MPa
True strain ε = ln(Lf/Lo) = ln(Ao/Af) = ln(4) = 1.386. However, it should be noted that these values
are associated with the necked portion of the test specimen.
3.19 A steel tensile specimen with starting gage length = 2.0 in and cross-sectional area = 0.5 in2 reaches a
maximum load of 37,000 lb. Its elongation at this point is 24%. Determine the true stress and true
strain at this maximum load.
Solution: Elongation = (L - Lo)/Lo = 0.24
L - Lo = 0.24 Lo
L = 1.24 Lo
A = Ao/1.24 = 0.8065 Ao
True stress σ = 37,000/0.8065(0.5) = 91,754 lb/in2
True strain ε = ln(1.24) = 0.215
Compression
3.20 A metal alloy has been tested in a tensile test with the following results for the flow curve parameters:
strength coefficient = 620.5 MPa and strain-hardening exponent = 0.26. The same metal is now tested
in a compression test in which the starting height of the specimen = 62.5 mm and its diameter = 25
mm. Assuming that the cross section increases uniformly, determine the load required to compress
the specimen to a height of (a) 50 mm and (b) 37.5 mm.
Solution: Starting volume of test specimen V = πhDo2/4 = 62.5π(25)2/4 = 30679.6 mm3.
(a) At h = 50 mm, ε = ln(62.5/50) = ln(1.25) = 0.223
Yf = 620.5(.223).26 = 420.1 MPa
A = V/L = 30679.6/50 = 613.6 mm2
F = 420.1(613.6) = 257,770 N
(b) At h = 37.5 mm, ε = ln(62.5/37.5) = ln(1.667) = 0.511
Yf = 620.5(0.511).26 = 521.1 MPa
A = V/L = 30679.6 /37.5 = 818.1 mm2
F = 521.1(818.1) = 426,312 N
3.21 The flow curve parameters for a certain stainless steel are strength coefficient = 1100 MPa and strain-
hardening exponent = 0.35. A cylindrical specimen of starting cross-section area = 1000 mm2 and
height = 75 mm is compressed to a height of 58 mm. Determine the force required to achieve this
compression, assuming that the cross-section increases uniformly.
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Solutions for Fundamentals of Modern Manufacturing, 3/e (published by Wiley) © MPGroover 2007
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Solutions for Fundamentals of Modern Manufacturing, 3/e (published by Wiley) © MPGroover 2007
yields a value of HB = 118. (a) Does the steel meet the specification on tensile strength? (b) Estimate
the yield strength of the material.
Solution: (a) TS = 500(BHN) = 500(118) = 59,000 lb/in2. This lies outside the specified range of
60,000 to 70,000 lb/in2. However, from a legal standpoint, it is unlikely that the batch can be
rejected on the basis of its measured Brinell hardness number without using an actual tensile test to
measure TS. The formula for converting from Brinell hardness number to tensile strength is only an
approximating equation.
(b) Based on Table 3.2 in the text, the ratio of Y to TS for low carbon steel = 25,000/45,000 =
0.555. Using this ratio, we can estimate the yield strength to be Y = 0.555(59,000) = 32,700 lb/in2.
Viscosity of Fluids
3.34 Two flat plates, separated by a space of 4 mm, are moving relative to each other at a velocity of 5
m/sec. The space between them is occupied by a fluid of unknown viscosity. The motion of the plates
is resisted by a shear stress of 10 Pa due to the viscosity of the fluid. Assuming that the velocity
gradient of the fluid is constant, determine the coefficient of viscosity of the fluid.
Solution: Shear rate = (5 m/s x 1000 mm/m)/(4 mm) = 1250 s-1
η = (10N/m2)/(1250 s-1) = 0.008 N-s/m2.
3.35 Two parallel surfaces, separated by a space of 0.5 in that is occupied by a fluid, are moving relative to
each other at a velocity of 25 in/sec. The motion is resisted by a shear stress of 0.3 lb/in2 due to the
viscosity of the fluid. If the velocity gradient in the space between the surfaces is constant, determine
the viscosity of the fluid.
Solution: Shear rate = (25 in/sec)/(0.5 in) = 50 sec-1
η = (0.3 lb/in2)/(50 sec-1) = 0.0006 lb-sec/in2.
3.36 A 125.0 mm diameter shaft rotates inside a stationary bushing whose inside diameter = 125.6 mm
and length = 50.0 mm. In the clearance between the shaft and the bushing is contained a lubricating
oil whose viscosity = 0.14 Pas. The shaft rotates at a velocity of 400 rev/min; this speed and the
action of the oil are sufficient to keep the shaft centered inside the bushing. Determine the magnitude
of the torque due to viscosity that acts to resist the rotation of the shaft.
Solution: Bushing internal bearing area A = π(125.6)2 x 50/4 = 19729.6 mm2 = 19729.2(10-6) m2
d = (125.6 - 125)/2 = 0.3 mm
v = (125π mm/rev)(400 rev/min)(1 min/60 sec) = 2618.0 mm/s
Shear rate = 2618/0.3 = 8726.6 s-1
τ = (0.14)(8726.6) = 1221.7 Pa = 1221.7 N/mm2
Force on surface between shaft and bushing = (1221.7 N/mm2)(19729.2(10-6)) = 24.1 N
Torque T = 24.1 N x 125/2 mm = 1506.4 N-mm = 1.506 N-m
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